Give reasons for the following:
(i) $Z n^{+ 2}$ salts are while but $C u^{2 +}$ salts are blue in colour.
(ii) Fluorine gives only one oxide but chlorine gives a series of oxides.

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Solution

(i) Zinc and copper are transition metals and their ions are generally coloured due to d-d transition of unpaired electron.
But in case of $Z n^{2 +} 3 d$ is completely filled and no unpaired electrons are available for transition. So light falling on these ions gets completely transmitted and their salts appear white. How ever ion $C u^{2 +} 3 d$ has $9$ electrons. The unpaired electron absorbs particular wavelengths in visible region of light and the transmitted light shows the complementary colour 'blue'.
(ii) Fluorine has only two orbits. In $2$ nd orbit it has $s^{2} p^{5}$ configuration with no vacant 'd' orbitals. So it can not form more than one oxide i.e., $O F_{2}$ .
Chlorine, on the other hand, has $3$ orbits in its atoms and $3 d$ is lying vacant in it. Therefore it can show variable covalency and gives a series of oxides. Like $C l_{2} O, C l O_{2}$ .

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